Question

What volume of a 15.0% by mass NaOH solution, which has a density of 1.116 g/mL, should be used to make 5.20 L of an NaOH solution with a pH of 10.0?
Express your answer to one significant figure and include the appropriate units.

Answer 1

Ans. #Step 1: Required [H₃O⁺] in the solution = 10^-pH = 10^-10 M

And, Required [OH^-] in the solution = 10^-14 / [H₃O⁺] = 10^-14 / 10^-10 = 10^-4 M

Since 1 mol NaOH yields 1 mol OH^-, the required concentration of NaOH must be equal to that of OH^-.

So, required [NaOH] in the solution = 10^-4 M

And, Required moles of NaOH = [NaOH] x Vol. of soln. in liters

                                                = 10^-4 M x 5.20 L = 5.20 x 10^-4 mol

# Step 2: Let the required volume of 15.0% stock NaOH solution be X mL.

Now,

            Mass of solution = Vol. taken x Density

                                                = X mL x 1.116 g mL^-1 = 1.116X g

Mass of NaOH taken = [NaOH] (w/w) x Mass of solution taken

                                                = 15.0% (w/v) x 1.116X g

                                                = (15.0 g /100 g) x 1.116X g

                                                = 0.1674X g

            Moles of NaOH taken = Mass / molar mass

                                                = 0.1674X g / 40.00 g mol^-1 = 0.004185X mol

#Step 3: From #Step 1, we have the required moles of NaOH to be 5.20 x 10^-4 mol.

Comparing the two values-

                        0.004185X mol = 5.20 x 10^-4 mol

                        Or, X = (5.20 x 10^-4) / 0.004185 = 0.1243

Hence, the required volume of stock NaOH solution = X mL = 0.1243 mL = 1.24 x 10^-1 mL